For the following equilibrium
`NH_(3)+H_(2)O hArr NH_(4)^(+)+OH^(-)`
calculate the equilibrium constant, if for the equilibrium,
`NH_(4)^(+)+H_(2)OhArr NH_(4)OH+H^(+)`
the equilibrium constant is `5.5xx10^(-10)`

Options

A. `1.8xx10^(-4)`
B. `1.8xx10^(-5)`
C. `1.8xx10^(-6)`
D. `1.8xx10^(-7)`

Answer

To calculate the equilibrium constant for the reaction of ammonia with water, we can follow these steps:

### Step 1: Write Down the Given Equilibrium Reaction and Constant
The equilibrium reaction we are interested in is:
\[ \text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]
We need to find the equilibrium constant \( K_1 \) for this reaction.

We are given another equilibrium reaction:
\[ \text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4\text{OH} + \text{H}^+ \]
with an equilibrium constant \( K_2 = 5.5 \times 10^{-10} \).

### Step 2: Write the Expression for the Given Equilibrium Constant
For the second reaction, the equilibrium constant \( K_2 \) can be expressed as:
\[ K_2 = \frac{[\text{NH}_4\text{OH}][\text{H}^+]}{[\text{NH}_4^+][\text{H}_2\text{O}]} \]

### Step 3: Write the Expression for the First Reaction
For the first reaction, the equilibrium constant \( K_1 \) can be expressed as:
\[ K_1 = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3][\text{H}_2\text{O}]} \]

### Step 4: Relate the Two Reactions
We can relate the two reactions by manipulating the second reaction. If we reverse the second reaction, we get:
\[ \text{NH}_4\text{OH} + \text{H}^+ \rightleftharpoons \text{NH}_4^+ + \text{H}_2\text{O} \]
The equilibrium constant for the reverse reaction \( K_{-2} \) is the reciprocal of \( K_2 \):
\[ K_{-2} = \frac{1}{K_2} = \frac{1}{5.5 \times 10^{-10}} \]

### Step 5: Combine the Two Reactions
Now, we can combine the reversed second reaction with the first reaction:
1. The reversed second reaction gives us \( \text{NH}_4^+ + \text{H}_2\text{O} \) on the left.
2. The first reaction gives us \( \text{NH}_3 + \text{H}_2\text{O} \) on the left.

When we combine these, we can cancel out \( \text{NH}_4^+ \) and \( \text{H}_2\text{O} \) to get:
\[ \text{NH}_3 + \text{H}^+ + \text{OH}^- \rightleftharpoons \text{NH}_4^+ \]

### Step 6: Write the Expression for the Combined Equilibrium Constant
The equilibrium constant for the combined reaction can be expressed as:
\[ K_1 = K_{-2} \times K_2 \]

### Step 7: Substitute the Values
Substituting the values:
\[ K_1 = \left(\frac{1}{5.5 \times 10^{-10}}\right) \times (10^{-14}) \]

### Step 8: Calculate \( K_1 \)
Calculating \( K_1 \):
\[ K_1 = \frac{10^{-14}}{5.5 \times 10^{-10}} = 1.818 \times 10^{-5} \]

### Step 9: Round Off the Result
Rounding off, we get:
\[ K_1 \approx 1.8 \times 10^{-5} \]

### Final Answer
Thus, the equilibrium constant for the reaction \( \text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \) is:
\[ K_1 = 1.8 \times 10^{-5} \]

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